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AS Physics P1 MCQs Preparation Thread.

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IVE POSTED THE SAME SHIT, ABOUT 5 TIMES NOW AND YOU WONT ANSWER, SO DONT TELL ME TO ASK NICELY OR SAY THIS OR THAT, CAUSE IT OBVIOUSLY DONT WORK

maybe because you don't have a catchy signature nor avatar

and so what i posted my questions 5 times before Sana101 answered them
 

Nibz

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lol i think Nibz edited it:p coz first he had written Q downwards.......whatever the case i still don't get it.....did u? If yes pleeasee do help:(
I edited it because it was a typing mistake. Don't show me your tongue.
Writing Q instead of P doesn't make any difference as long as the explanation is right.
 

Nibz

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would you care to elaborate what you explained before?

There is nothing to elaborate, dear.
From trough to crest, particles are in downward motion; From crest to trough, they are in upward motion.
At max point, they are momentarily at rest.
 
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There is nothing to elaborate, dear.
From trough to crest, particles are in downward motion; From crest to trough, they are in upward motion.
At max point, they are momentarily at rest.
hmmmm starting to make sense:) thanx
 
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Since all questions have been explained apart from 15 so
For q15 Efficency=Useful Output energy/Input Energy multiplied by 100
Now Input energy=Force*Distance=Fs
For output energy if u read the question carefully the car is moving with constant speed v so it is not gaining kinetic energy.It is only gaining Potential energy.Now u have to calculate height.Whenever u need to calculate h and distance of slope and angle of slope with horizontal is given,use the formula
sin alpha=h/s so h=s sin alpha
Gain in P.E=m*g*s sinalpha
m*g*s sinalpha divided by Fs giving u D as answer
 
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hay can any1 help me with this pretty plz i kinda suck @ physics:(

A motorist travelling at 10ms–1 can bring his car to rest in a braking distance of 10m. In what distance could he bring the car to rest from a speed of 30 m s–1 using the same braking force?
A 17m B 30m C 52m D 90m

the marking scheme ans is D
so any 1 goody good out there mynd helping me out plzzzzzzzzzzzzzz:p
 
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Okay, here it is again :
Questions: 10, 11 , 21, 24, 26, 27, 31, 33 ,36

PLEASE HELP URGENTLY!!!! AND EXPLAIN AS WELL, I CAN GET THE ANSWERS FROM A MARK SCHEME I NEED EXPLANATIONS!! THANK YOU GOD BLESS

And i uploaded the question paper too.
 

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Nibz

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Okay, here it is again :
Questions: 10, 11 , 21, 24, 26, 27, 31, 33 ,36

PLEASE HELP URGENTLY!!!! AND EXPLAIN AS WELL, I CAN GET THE ANSWERS FROM A MARK SCHEME I NEED EXPLANATIONS!! THANK YOU GOD BLESS

And i uploaded the question paper too.

Q10.

mu1 - mu2 = (m+m)v
60m - 40m = 2mv
20m/2m = v
10 = v


Q11.
Net force = m x a
10 - 4 = 2a
6/2 = a
3 = a

Q24.
f = v/lambda => wavelength and frequency have an inverse proportionality. Increasing f would decrease the wavelength.
With shorter wavelength, diffraction is less pronounced, i.e more closer to diagram 2.

Q31.
R = ρL/A
R . A /L = ρ
Ω m^2 /m = ρ
Ω m = ρ
 
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